On a space shuttle, LiOH is used to absorb exhaled $CO_{2}$ from breathing air to form $LiHCO_{3}$ .
$LiOH(s) + CO_{2}(g) → LiHCO_{3}(s)$
What is the percent yield of $LiHCO_{3}$ for the reaction if 50.0 g of LiOH gives 72.8 g of $LiHCO_{3}$ ?

Question

On a space shuttle, LiOH is used to absorb exhaled [latex] CO_{2}[/latex] from breathing air to form [latex] LiHCO_{3}[/latex].

[latex] LiOH(s) + CO_{2}(g) → LiHCO_{3}(s)[/latex]

What is the percent yield of [latex] LiHCO_{3}[/latex] for the reaction if 50.0 g of LiOH gives 72.8 g of [latex] LiHCO_{3}[/latex]?

Accepted Answer

STEP 1 State the given and needed quantities (moles).

ANALYZE THE PROBLEM	Given	Need	Connect
	50.0 g of LiOH (reactant), 72.8 g of [latex] LiHCO_{3}[/latex] (actual product)	percent yield of [latex] LiHCO_{3}[/latex]	molar masses, mole-mole factor , percent yield expression
	Equation
	[latex] LiOH(s) + CO_{2}(g) → LiHCO_{3}(s) [/latex]

STEP 2 Write a plan to calculate the theoretical yield and the percent yield.
Calculation of theoretical yield:

[latex] grams of LiOH \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → moles of LiOH \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}} → moles of LiHCO_{3} \boxed{\begin{array}{l} ext{Molar}\ ext{Mass}\end{array}} → \underset{Theoretical yield}{grams of LiHCO_{3}}[/latex]

Calculation of percent yield:

Percent yield(%)=[latex] \frac{acntal yield of LiHCO_{3}}{theoretical yield of LiHCO_{3}} imes 100[/latex]%

STEP 3 Use coefficients to write mole-mole factors; write molar mass factors.

[latex] \begin{array}{r c}\boxed{\begin{matrix} 1 mole of LiOH = 23.95 g of LiOH \\frac{23.95 g LiOH}{1 mole LiOH} ext{ and } \frac{1 mole LiOH}{23.95 g LiOH} \end{matrix}} & \boxed{\begin{matrix} 1 mole of LiHCO_{3} = 67.96 g of LiHCO_{3} \ \frac{67.96 g LiHCO_{3}}{ 1 mole LiHCO_{3}} ext{ and }\frac{1 mole LiHCO_{3}}{67.96 g LiHCO_{3}} \end{matrix}}\end{array} [/latex]

[latex] \begin{array}{r c}\boxed{\begin{matrix} 1 mole of LiHCO_{3}= 1 mole of LiOH \\frac{1 mole LiHCO_{3}}{1 mole LiOH} ext{ and } \frac{1 mole LiOH}{1 mole of LiHCO_{3}} \end{matrix}}\end{array} [/latex]

STEP 4 Calculate the percent yield by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%.
Calculation of theoretical yield:

[latex] \underset{Three SFs }{50.0 \cancel{g LiOH}} imes \overset{Exact}{\underset{Four SFs }{\boxed{\frac{1 \cancel{mole LiOH}}{23.95 \cancel{ g LiOH}} }}} imes \overset{Exact}{\underset{Exact }{\boxed{\frac{1 \cancel{mole LiHCO_{3}}}{1 \cancel{ mole LiOH}} }}} imes \overset{Four SFs}{\underset{Exact }{\boxed{\frac{67.96 g LiHCO_{3}}{1 \cancel{ mole LiHCO_{3}}} }}} = \underset{Three Sfs }{142 g of LiHCO_{3}}[/latex]

Calculation of percent yield:

[latex] \frac{acntal yield (given) }{theoretical yield (calculated)} imes 100 [/latex]% = [latex] \overset{Three Sfs}{\underset{Three Sfs}{\frac{72.8 \cancel{g LiHCO_{3}}}{142 \cancel{g LiHCO_{3}}}}} imes 100 [/latex]% =[latex]\underset{Three Sfs }{51.3}[/latex]%

Question 7.15: On a space shuttle, LiOH is used to absorb exhaled CO2 from ...

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